1. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
Create equations and inequalities in one variable and use them to solve problems.
Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational, absolute, and exponential functions.

2. MAFS.912.A-CED.1.1 What You Need To Know... Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
What You Need To Know...
1. The benchmark will be assessed using Drag and Drop Response, Equation Response, Hot Spot Response, Multiple Choice Response or Open Response.
2. Items may include equations from linear and quadratic functions and simple rational, absolute and exponential functions.
3. Items may require students to rearrange formulas to highlight a particular variable.
4. Items may require students to recognize equivalent expressions.

3. MAFS.912.A-CED.1.1 Mrs. Rivas Standard Example 1
ISCHS
Standard
MAFS.912.A-CED.1.1
Ronald can deliver papers in his neighborhood 8 hours faster than his little brother Jimmy can.  If they work together, they can complete the job in 3 hours.  Solve for how long it would take for Jimmy to complete this job on his own.  Using complete sentences, explain each step of your work.
Example 1
Jimmy’s rate is 1 𝒙 hours or 𝟏 𝒙
Ron’s rate is 1 (𝒙−𝟖) or 𝟏 𝒙 + 𝟖
Together their rate is 1 3 hours or 1 3
𝟏 𝒙 + 𝟏 𝑥 − 𝟖 = 𝟏 𝟑
𝑳𝑪𝑫=𝟑𝒙(𝒙−𝟖)
𝟑(𝒙−𝟖)+𝟑𝒙=𝒙(𝒙−𝟖)
𝟑𝒙(𝒙−𝟖) 𝟏 𝒙 + 𝟏 𝑥 − 𝟖 = 𝟏 𝟑
𝟑𝒙−𝟐𝟒+𝟑𝒙= 𝒙 𝟐 −𝟖𝒙
𝟔𝒙−𝟐𝟒= 𝒙 𝟐 −𝟖𝒙
𝟎= 𝒙 𝟐 −𝟏𝟒𝒙+𝟐𝟒
𝟑𝒙(𝒙−𝟖) 𝒙 + 𝟑𝒙(𝒙−𝟖) 𝑥 + 𝟖 = 𝟑𝒙(𝒙−𝟖) 𝟑
𝟎=(𝒙−𝟏𝟐)(𝒙−𝟐)
𝒙 = 𝟏𝟐 or x = 2 but when you substitute in the original equation Jimmy can’t be equal to 2 because then Ron would be completing the job in negative hours. So Jimmy’s rate is 12 and Ron’s is 4.

4. MAFS.912.A-CED.1.1 Mrs. Rivas Standard Example 2
ISCHS
Standard
MAFS.912.A-CED.1.1
The high school gym holds a maximum of 1120 people. The drama club is performing a skit in the gym this Friday. There are 40 students in drama club. Write and solve an inequality to determine how many guests each student can invite to the skit.
Example 2
Since the gym will hold a maximum of 1120 people, the number of guests and students must be less than or equal to 1120.
Let g represent the number of guests.
𝟒𝟎+𝟒𝟎𝒈 ≤𝟏𝟏𝟐𝟎
40 students + 40 times the number of guest per student
−𝟒𝟎
−𝟒𝟎
𝟒𝟎𝒈 ≤𝟏𝟎𝟖𝟎 𝟒𝟎 𝟒𝟎
𝒈 ≤𝟐𝟕
Each student can invite a maximum of 27 guests

5. Rate of growth or decay (percentage)
Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
1. John is buying a car for $8,000. The value of the car will decrease by 5% each year. Which equation can he use to predict the value of the car after 3 years?
Amount after 𝒕 periods
A. 𝑦=8,
B. 𝑦=8,000 1−0.5 3
𝑨(𝒕)=𝒂(𝟏+𝒓) 𝒕
C. 𝑦=8,000 1−
D. 𝑦=8,
Number of time period
Initial
amount
Rate of growth or decay (percentage)

6. Key word:
Decrease
Key word:
Decrease
Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
1. John is buying a car for $8,000. The value of the car will decrease by 5% each year. Which equation can he use to predict the value of the car after 3 years?
1. John is buying a car for $8,000. The value of the car will decrease by 5% each year. Which equation can he use to predict the value of the car after 3 years?
$8,000
decrease by 5%
each year
Which equation can he
use to predict the value of the car after 3
years
A. 𝑦=8,
𝑨(𝒕)=𝒂(𝟏+𝒓) 𝒕
𝒚=𝟖,𝟎𝟎𝟎(𝟏−𝟎.𝟎𝟓) 𝒕
B. 𝑦=8,000 1−0.5 3
𝒚=𝟖,𝟎𝟎𝟎(𝟏−𝟎.𝟎𝟓) 𝟑
C. 𝑦=8,000 1−
D. 𝑦=8,
1) What are they asking me?
2) What are they giving me?

7. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
2. After sitting out of the refrigerator for a while, a turkey at room temperature (68°F) is placed into an oven at 8 a.m., when the oven temperature is 325°F. Newton’s Law of Heating explains that the temperature of the turkey will increase proportionally to the difference between the temperature of the turkey and the temperature of the oven, as given by the formula below:
𝑻= 𝑻 𝒔 + 𝑻 𝟎 + 𝑻 𝒔 𝒆 −𝒌𝒕
𝑻 𝒔 : the temperature surrounding the object.
𝑻 𝟎 : the initial temperature of the object.
𝒕: the time in hours.
𝑻: the temperature of the object after 𝑡 hours
𝒌: decay constant.
𝑻 𝒔 : 𝟑𝟐𝟓℉
𝑻 𝟎 : 𝟔𝟖℉
𝒕: 𝟐 𝒉𝒐𝒖𝒓𝒔
𝑻: 𝟏𝟎𝟎℉
The turkey reaches the temperature of approximately 100° F after 2 hours. Find the value of 𝑘, to the nearest thousandth, and write an equation to determine the temperature of the turkey after 𝑡 hours. Determine the Fahrenheit temperature of the turkey, to the nearest degree, at 3 p.m.
𝒌≈𝟎.𝟎𝟔𝟔 𝒂𝒏𝒅 𝑻≈𝟏𝟔𝟑

8. Key Concept Natural Logarithmic Function
Mrs. Rivas
ISCHS
Standard
Key Concept Natural Logarithmic Function
𝒚= 𝒆 𝒙  𝒙= 𝐥𝐨𝐠 𝒆 𝒚
The natural logarithmic function is the inverse
𝒚= 𝒆 𝒙  ln𝒚=ln⁡ 𝒆 𝒙  ln𝒚= 𝒆 𝒙
MAFS.912.A-CED.1.1
2. After sitting out of the refrigerator for a while, a turkey at room temperature (68°F) is placed into an oven at 8 a.m., when the oven temperature is 325°F. Newton’s Law of Heating explains that the temperature of the turkey will increase proportionally to the difference between the temperature of the turkey and the temperature of the oven, as given by the formula below:
Find the value of k?
𝑻 𝒔 : 𝟑𝟐𝟓℉
𝑻= 𝑻 𝒔 + 𝑻 𝟎 − 𝑻 𝒔 𝒆 −𝒌𝒕
𝑻 𝟎 : 𝟔𝟖℉
𝟏𝟎𝟎=𝟑𝟐𝟓+ 𝟔𝟖−𝟑𝟐𝟓 𝒆 −𝒌(𝟐)
𝒕: 𝟐 𝒉𝒐𝒖𝒓𝒔
𝟏𝟎𝟎=𝟑𝟐𝟓−𝟐𝟓𝟕 𝒆 −𝒌(𝟐)
𝑻: 𝟏𝟎𝟎℉
−𝟐𝟐𝟓=−𝟐𝟓𝟕 𝒆 −𝒌(𝟐)
𝟎.𝟖𝟕𝟓𝟓= 𝒆 −𝒌(𝟐)
−𝟎.𝟏𝟑𝟐𝟗𝟔=−𝟐𝒌
ln(𝟎.𝟖𝟕𝟓𝟓)=ln 𝒆 −𝒌(𝟐)
𝒌≈𝟎.𝟎𝟔𝟔
ln 𝟎.𝟖𝟕𝟓𝟓 =−𝟐𝒌

9. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
2. After sitting out of the refrigerator for a while, a turkey at room temperature (68°F) is placed into an oven at 8 a.m., when the oven temperature is 325°F. Newton’s Law of Heating explains that the temperature of the turkey will increase proportionally to the difference between the temperature of the turkey and the temperature of the oven, as given by the formula below:
Negative exponent
Determine the Fahrenheit temperature at 3 PM.
𝑻=𝟑𝟐𝟓+ 𝟔𝟖−𝟑𝟐𝟓 𝒆 −(𝟎.𝟎𝟔𝟔)(𝟕)
𝑻 𝒔 : 𝟑𝟐𝟓℉
𝑻 𝟎 : 𝟔𝟖℉
𝑻=𝟑𝟐𝟓−𝟐𝟓𝟕 𝒆 −(𝟎.𝟎𝟔𝟔)(𝟕)
𝑻: 𝟏𝟎𝟎℉
𝑻=𝟑𝟐𝟓− 𝟐𝟓𝟕 𝒆 (𝟎.𝟎𝟔𝟔)(𝟕)
𝒌≈𝟎.𝟎𝟔𝟔
𝒕: 𝟕 𝒉𝒐𝒖𝒓𝒔
𝑻=𝟑𝟐𝟓− 𝟐𝟓𝟕 𝒆 (𝟎.𝟒𝟔𝟐)
𝑻≈𝟏𝟔𝟑° 𝑭
8 a.m. to 3p.m are 7 hours (𝒕)
𝑻=𝟑𝟐𝟓−𝟏𝟔𝟏.𝟗𝟏𝟓𝟕

10. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.4
3. The period for a pendulum to complete one swing is 𝑡, the time in seconds. The period can be approximated by the formula t=2π l , where 𝑙 is the length of the pendulum in meters. If the period of a pendulum is 2.5 seconds, which is closest to the length of the pendulum?
1) What are they asking me?
A meters
B meters
C meters
D meters
𝒍=Length of the pendulum
2) What are they giving me?
𝒕: period for a pendulum = 𝟐.𝟓 sec.
𝒕=𝟐𝝅 𝒍 𝟗.𝟖𝟏
𝟐.𝟓=𝟐(𝟑.𝟏𝟒) 𝒍 𝟗.𝟖𝟏
𝟐.𝟓=𝟔.𝟐𝟖 𝒍 𝟗.𝟖𝟏
𝟎.𝟑𝟗𝟖𝟏 𝟐 = 𝒍 𝟗.𝟖𝟏 𝟐
𝟎.𝟑𝟗𝟖𝟏= 𝒍 𝟗.𝟖𝟏
𝟎.𝟏𝟓𝟖𝟓= 𝒍 𝟗.𝟖𝟏
𝒍≈𝟏.𝟓𝟓𝟒𝟕

11. MAFS.912.A-CED.1.1 Mrs. Rivas 4. What is the solution to 5𝑥+6 +3=7?
ISCHS
Standard
MAFS.912.A-CED.1.1
4. What is the solution to 5𝑥+6 +3=7?
A. 𝑥= 4 5
B. 𝑥=2
C. 𝑥= 34 5
D. 𝑥=8
𝟓𝒙+𝟔 +𝟑=𝟕
𝟓𝒙+𝟔 =𝟕−𝟑
𝟓𝒙+𝟔 =𝟒
𝟓𝒙+𝟔 𝟐 = 𝟒 𝟐
𝟓𝒙+𝟔=𝟏𝟔
𝟓𝒙=𝟏𝟔−𝟔
𝟓𝒙=𝟏𝟎
𝒙=𝟐

12. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
5. What nonzero value of 𝑥 is a solution to the following equation?
A. 𝑥= 27 14
B. 𝑥= 17 14
C. 𝑥= 13 14
D. 𝑥= 5 14
𝒙+𝟐 𝒙 + 𝒙−𝟔 𝟑𝒙 = 𝟐𝒙+𝟗 𝟓𝒙
𝑳𝑪𝑫=𝟑∙𝟓∙𝒙
𝟑∙𝟓∙𝒙 𝒙+𝟐 𝒙 + 𝟑∙𝟓∙𝒙 𝒙−𝟔 𝟑𝒙 = 𝟑∙𝟓∙𝒙 𝟐𝒙+𝟗 𝟓𝒙
𝟏𝟓 𝒙+𝟐 +𝟓 𝒙−𝟔 =𝟑(𝟐𝒙+𝟗)
𝟏𝟓𝒙+𝟑𝟎+𝟓𝒙−𝟑𝟎=𝟔𝒙+𝟐𝟕
𝟐𝟎𝒙=𝟔𝒙+𝟐𝟕
𝟏𝟒𝒙=𝟐𝟕
𝒙= 𝟐𝟕 𝟏𝟒

13. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
6. Solve algebraically for all values of 𝑥: 𝑥+5 +𝑥=7
𝒙=𝟒 𝒂𝒏𝒅 𝟏𝟏; 11 is an extraneous solution
𝒙−𝟏𝟏=𝟎
𝒙−𝟒=𝟎
𝒙+𝟓 +𝒙=𝟕
𝒙=𝟏𝟏
𝒙=𝟒
𝒙+𝟓 =𝟕−𝒙
𝒙+𝟓 𝟐 = 𝟕−𝒙 𝟐
𝑪𝒉𝒆𝒄𝒌
𝟏𝟏+𝟓 +𝟏𝟏≠𝟕
𝒙+𝟓= 𝒙 𝟐 −𝟏𝟒𝒙+𝟒𝟗
𝟎= 𝒙 𝟐 −𝟏𝟓𝒙+𝟒𝟒
𝟒+𝟓 +𝟒=𝟕 𝒙
−𝟏𝟏
𝟑+𝟒=𝟕 𝒙 −𝟒
𝟕=𝟕
𝒙−𝟏𝟏 𝒙−𝟒 =𝟎

14. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
7. What extraneous solution arises when the equation 𝑥+3 =2𝑥 is solved for 𝑥 by first squaring both sides of the equation?
Enter your answer in the box.
𝟒𝒙+𝟑=𝟎
𝒙−𝟏=𝟎
−𝟎.𝟕𝟓
𝒙=− 𝟑 𝟒
𝒙=− 𝟑 𝟒
𝒙=𝟏
𝒙+𝟑 =𝟐𝒙
𝒙+𝟑 𝟐 = 𝟐𝒙 𝟐
𝑪𝒉𝒆𝒄𝒌
− 𝟑 𝟒 +𝟑 =𝟐 − 𝟑 𝟒
𝒙+𝟑= 𝟒𝒙 𝟐
𝟑 𝟐 ≠− 𝟑 𝟐
𝟎= 𝟒𝒙 𝟐 −𝒙−𝟑 𝟒𝒙 −𝟑 𝒙 −𝟏
𝟏+𝟑 =𝟐 𝟏
𝟒 =𝟐
𝟒𝒙−𝟑 𝒙−𝟏 =𝟎
𝟐=𝟐

15. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
8. Determine the solution(s) of the equation.
𝟐 𝒎 𝟐 +𝟑𝒙−𝟓 𝒎 𝟐 +𝟒𝒎−𝟓 =𝟒
A. −5
B. − 15 2
C. − 5 2
D. 1
(𝒙−𝟏)(𝟐𝒙+𝟓) (𝒙−𝟏)(𝒙+𝟓) =𝟒
(𝒙+𝟓)
(𝟐𝒙+𝟓) (𝒙+𝟓) =𝟒
(𝒙+𝟓)
𝟐𝒙+𝟓=𝟒𝒙+𝟐𝟎
𝟐𝒙=𝟒𝒙+𝟏𝟓
−2𝑥=15
𝒙=− 𝟏𝟓 𝟐

16. Mrs. Rivas
ISCHS
Standard
MAFS.912.A-CED.1.1
9. The speed of a tidal wave, 𝑠, in hundreds of miles per hour, can be modeled by the equation s=√t−2t+6, where 𝑡 represents the time from its origin in hours. Algebraically determine the time when s= 0.
𝟎= 𝒕 −𝟐𝒕+𝟔
How much faster was the tidal wave traveling after 1 hour than 3 hours, to the nearest mile per hour? Justify your answer.
(𝒕−𝟒)(𝟒𝒕−𝟗)=𝟎
𝒕=𝟒
𝟐𝒕−𝟔= 𝒕
𝒕−𝟒=𝟎
𝟒𝒕−𝟗=𝟎
𝟐𝒕−𝟔 𝟐 = 𝒕 𝟐
𝒕=𝟒
𝒕= 𝟗 𝟒
𝟒𝒕 𝟐 −𝟐𝟒𝒕+𝟑𝟔=𝒕
𝟒𝒕 𝟐 −𝟐𝟓𝒕+𝟑𝟔=𝟎
How much faster was the tidal wave traveling after 1 hour than 3 hours, to the nearest mile per hour? Justify your answer.
𝒔= 𝟏 −𝟐 𝟏 +𝟔=𝟓
𝒔= 𝟑 −𝟐 𝟑 +𝟔=𝟏.𝟕𝟑𝟐𝟎𝟓
The tidal was traveling 3.27 miles per hour faster